Implicit multivariable derivative of a spherical cap

1. Jun 7, 2014

babagoslow

1. The problem statement, all variables and given/known data

Consider a spherical cap, for which the surface area and volume is

$$A(a,h) = \pi(a^2 +h^2)$$
$$V(a,h) = \frac{\pi h}{6}(3a^2 +h^2)$$

What would the aspect ratio $dA/dV$ be?

3. The attempt at a solution

Clearly we would have

$$dA = 2\pi a da + 2\pi h dh$$
$$dV = \pi ha da + \frac{\pi}{2}(a^2+h^2) dh$$

but what would the chain rule look like? Surely, it should have more terms than just

$$\frac{dA}{dV } = \frac{\partial A}{\partial a} \frac{\partial a}{\partial V} + \frac{\partial A}{\partial h}\frac{\partial h}{\partial V}$$

, right?

2. Jun 9, 2014

HallsofIvy

Staff Emeritus
Why do you think it should have more terms? a and h are the only variables.

3. Jun 9, 2014

babagoslow

If $dV, dA$ were just functions of one variable, it would be straightforward. But here there are two variables, and perhaps I would have anticipated a chain rule of the form

$$\frac{dA}{dV} = \frac{\partial A}{\partial a}\left( \frac{\partial a}{\partial V} + \frac{\partial h}{\partial V} \right) + \frac{\partial V}{\partial a}\left( \frac{\partial a}{\partial A} + \frac{\partial h}{\partial A} \right)$$

but I'm just guessing.

4. Jun 9, 2014

pasmith

On dimensional grounds, I would expect $$A = V^{2/3} f\left(\frac ah\right)$$ for some function $f$ which can be determined from the formulae for $A$ and $V$. Thus $A$ is a function of both $V$ and $a/h$.

5. Jun 9, 2014

HallsofIvy

Staff Emeritus
The second part has V "in the numerator" that can't be right! "On dimensional grounds", as pasmith puts it, V has units of "length cubed", A has units of "length squared, and h and a have units of "length". $\partial V/\partial a$ has units of "length squared", and $\partial a/\partial A$ and $\partial h/\partial A$ have units of "1 over length" so the second term has units of length. But $dA/dV$ has units of "1 over length".

6. Jun 11, 2014

babagoslow

Thanks for the advice HallsofIvy and pasmith!

Pasmith, yes, you were right, I was able to evaluate the derivative by expressing the functions in terms of $f(h/a)$. Thanks for the insight!

7. Dec 19, 2014

babagoslow

Dear Pasmith, I have been looking at this problem again and in fact I don't think it is possible to find the derivative $dA/dV$ like you mentioned.

Let $x\equiv a/h.$
$A^3 = V^2 f(a/h) = V^2 f(x)$
$3A^2 dA = 2V dV f(x) + V^2 \frac{df}{dx} dx$

The complication comes in when you are trying to solve for $dA/dV$ and you get

$\frac{dA}{dV} = ... + \frac{V^2}{3A^2} \frac{df}{dx} \frac{dx}{dV}.$

But note that $V=\frac{\pi h}{6}(3a^2+h^2)$. Since the powers of $h, a$ are different, there is no way to evaluate the derivative $dx/dV$ without leaving behind some residual terms of $a, h$. The goal in the first place was to evaluate the derivative solely in $x\equiv a/h$. Therefore we can't parameterise the equation like you said here.

Can you help, Pasmith, or anybody? This problem has been bothering me for a while.

8. Dec 19, 2014

PeroK

A simpler approach is to use the relationship between $a$ and $h$ to express $A$ and $V$ in terms of $h$ and the radius of the sphere. That reduces it to a single variable problem.

9. Dec 19, 2014

babagoslow

$a , h$ are also related by a contact angle $\theta$. But you can't express a spherical cap uniquely in just that one variable.

I guess I don't really understand what you meant. From what I understand a spherical cap is uniquely described by at least two parameters - one possible combination is the the base width and height $a,h$, and another combination is the radius of curvature and angle $r,\theta$.

10. Dec 19, 2014

PeroK

I guess from what you said that this is not a spherical cap on a given sphere?

11. Dec 19, 2014

babagoslow

Yes, that's right, the cap is not on a given sphere. There are two parameters.

12. Dec 19, 2014

PeroK

So, are you not trying to calculate $\frac{\partial A}{\partial V}$ ?

13. Dec 19, 2014

babagoslow

That is what I am calculating. As we know, for a sphere, $V(r), A(r)$ and it's easy to calculate $\frac{dA}{dV} = \frac{dA}{dr} \frac{dr}{dV}$. But what happens when these two quantities are functions of more than one variable, ie, $V(a,h), A(a,h)$? Then how would we calculate $\partial A/\partial V$?

14. Dec 19, 2014

PeroK

I think your problem is that you also need to specify which variable is constant for your partial derivative. For example, you could express A as a function of V and h, say:

$A = f(V, h)$

And then you could calculate $\partial A/\partial V = \partial f/\partial V$ where h is held constant.

But, you could also find:

$A = g(V, a)$

And then you could calculate $\partial A/\partial V= \partial g/\partial V$ where a is held constant. And these would be different functions with different numerical values.

Or, you could hold $a/h$ constant etc.

That's why I assumed it was a fixed sphere, when you have V as a function of A with no other independent variables and can find $\frac{dA}{dV}$

15. Dec 19, 2014

Ray Vickson

Your question is not well-defined (as many others have also indicated). However, it is not very difficult to extract answers to more pointed questions. We have
$$\Delta A = 2 \pi a \, \Delta a + 2 \pi h \, \Delta h, \\ \Delta V = \pi h a \, \Delta a + \frac{\pi}{2}(a^2+h^2) \, \Delta h \\ \text{so}\\ \frac{\Delta A}{\Delta V} = \frac{2 \pi a \, \Delta a + 2 \pi h \, \Delta h}{\pi h a \, \Delta a + (\pi /2)(a^2+h^2) \, \Delta h}$$
This implies the following.
(1) If $a$ is a function of $h$ we get
$$\frac{dA}{dV} = \frac{2 \pi a (da/dh) + 2 \pi h}{\pi h a (da/dh) + (\pi/2) (a^2 + h^2)}$$
(2) If $h$ is a function of $a$ we get
$$\frac{dA}{dV} = \frac{2 \pi a+ 2 \pi h (dh/da)}{\pi h a + (\pi/2)(a^2+h^2) (dh/da)}$$
(3) If $h$ is a constant, the derivative $(dA/dV)_h$ --which is Thermodynamics-style notation, indicating that $h$ is held constant--has the form
$$\left( \frac{d A}{d V} \right)_{h} = \frac{ 2 \pi a}{\pi h a}.$$
(4) If $a$ is a constant, the derivative $(dA/dV)_a$ is
$$\left( \frac{dA}{dV}\right)_a = \frac{2 \pi h}{(\pi/2)(a^2+h^2)}$$

You get many different answers, depending on what you mean by the question to begin with.